So special

Number Theory Level 2

A three-digit number $N$ has first digit $a (\neq 0)$ , second digit $b$ and third digit $c$ . $N = b(10c + b)$ where $b$ and $(10c + b)$ are primes. Find $N$ .

The answer is 679.

7 solutions

Daniel Liu
Jun 21, 2015

Note that $b^2\equiv c\pmod{10}$ . Given that $b$ and $10c+b$ are prime, the possibilities are $b=3\text{ or } 7$ , but $93$ is not prime so $b=7$ . Thus $c=9$ and we calculate $7\times 97=\boxed{679}$

i haven't learned modular maths yet(i only know mod is the remainder of a division), but where do you get the rule b^2 is equivalent/congruence/etc c (mod 10)?

Andy Andy - 5 years, 7 months ago

Andy, what Daniel wrote:

$b^2 \equiv c \pmod{10}$

is just a fancy way of saying

"b*b will have a units digit of c."

That's because of two specific given criteria. First, N is given to be expressed abc.

And then the b(10c + b) part. See that the units digit is b*b?

If I were to write out my entire solution, I would have started with considering b as an element of {2,3,5,7}, i.e. all the single digit primes.

b = 2 is quickly eliminated because (10c + b) will only be a prime when c = 0. But the product b(10c + b) fails to be a three digit number. b(10c + b) = 4

For b = 3, note that 3(10c + 3) will have a 9 for the units digit, and that number is c.

But 3(10*9 + 3) = 279. Something is wrong. b is no longer 3. It is 7 when multiplied out. b = 3 is eliminated.

For b = 5, (10c + 5) will only be prime when c = 0. Like b = 2, we won't get a three digit number.

That leaves b = 7. 7(10c + 7) will end in a 9. And that 9 is c.

7(97) = (boxed{679}

This is exactly the same as Daniel's proof, I just need a few more steps to see it.

Ken Hodson - 5 years, 7 months ago

I can't understand how have you used $b^2\equiv c\pmod{10}$ ? Please elucidate your solution..

Prokash Shakkhar - 4 years, 6 months ago

Since N = abc and N (mod 10) is c, but N = b(10c+b) = 10cb + b^2. 10cb (mod 10) is 0 so b^2 (mod 10) must be c

Evan Fisher - 3 years, 5 months ago

I would like to bring in your notice that 3 and 73 also satisfy the above systems.😨

Aaryan Maheshwari - 3 years, 11 months ago

Nope. I don't think so. I think that 679, Is an unique solution.Maybe you messed up your calculations.

Shishir Shahi - 3 years, 11 months ago

@Shishir Shahi You can put b=3, c=7. Then 10c+b=70+3=73, which is prime.

Aaryan Maheshwari - 3 years, 11 months ago

@Aaryan Maheshwari – But, if you apply the values 10c+b=73,which is prime. But my point is that if you multiply this equation with b : (10c+b)(b)=73*3=219.Thus, if you see the question ten's digit of the number is b=3 but here the tens digit is 1. Thus you messed up.:)

Shishir Shahi - 3 years, 11 months ago

How exactly is it 'given' that b and 10c+b are prime?

I think a lot of these questions are more then a little light on definitions. You can't assume a variable in a problem is prime just because the question resides in a chapter about prime numbers.

Evelien Heerens - 2 years, 8 months ago

Malc Knight
May 13, 2017

b can be 2, 3, 5 or 7. But as10c+b must be also be prime, b can only be 3 or 7. As 2 and 5 are factors of 10, neither can be b. N=b(10c+b)=10cb+b^2. C is the last digit of b^2 which must be 9. 10c+b=97 as the alternative of 93 Is not prime. Therefore N=7*97=679

N=abc=b(10c+b) and both (10c+b) and b are prime b=3 c=5 10c+b=53 prime number b= 3 prime number N=3(10 5+3)=159 * It was never mentioned that N must be prime

mod dere - 3 years, 1 month ago

You will find that 10(9)+5 is not prime. The issue is that if you select b=3 and c=5, then the second and third digits must be 3 and 5

Michael Tamajong - 2 years, 3 months ago

b=5 can also be excluded, because the term (10c+b) is only prime for c=0 and the product b*(10c+b) will never result in a 3-digit number.

a t - 2 months, 4 weeks ago

Suvaditya Sur
Jun 11, 2016

b can be 2, 3, 5, 7 (bcoz b is a prime). But cb is prime => b can be 3 or 7. Enumerating all possible cb's = [13, 23, 43, 53, 73, 83, 17, 37, 47, 67, 97] wee se that 97 * 7 = 679 = 6cb. Thus 679 is the answer

Also. Since b (10c+b), being that b can only be 3 or 7, b^2 unit digit must be 9. Therefore c must be 9 thus b=7

Wellington Chen - 4 years, 9 months ago

K T
Mar 15, 2019

100a+10b+c=N=b(10c+b)

If b is 2 or 5, 10c+b will not be prime. If b=3, then b(10c+b) will end on a 9, so c must be 9. But 93×3=279, the second digit of which is b=7, but b cannot be 3 and 7 in the same time. If b=7 then c must be 9. 7×97=679, which is consistent.

Albaraa Alsharif
Mar 7, 2019

Scrub Lord
Apr 5, 2017

b is either 3 or 7. That gives us two linear diophantine equations:

100a - 29c +21 = 0 for b=3,

100a - 69c + 21 = 0 for b = 7.

In the first case, the solution is a = 504 - 49k and c = 1029 -100k. We can see that there is no such k so that a and c would be integers and a, c < 10. Therefore, b = 7.

The solution for the second equation are a = 420 - 69k and c = 609 -100k. Put in 6 for k, and we get a = 6 and c = 9. N = 679.

Edmund Zobrist
Jan 14, 2020

As 10c+b is prime then only possible choices for b are 3 7 Now b2 congruent c modulo 10 . Checking 3 and 9 . Then for b=3 ,c=-1 So b=7 and so folows

So special

The answer is 679.

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